Assume firstly that \(p\)  and \(q\)  verify conditions (REF ). So, by [1]}, the 2-class group of \(L_{pq}\)  is of type \((2, 2)\) . Under this assumption, \(h_2(p)=h_2(q)=h_2(-2)=1\) ,
\(h_2(-2p)=h_2(pq)=2\)  and \(h_2(-2pq)=4\)  (cf.[2]}, [3]}).
We claim that the class number of \(k^{\prime }=\mathbb {Q}(\sqrt{p}, \sqrt{q})\)  is odd. In fact, we have
\(h_2(p)=h_2(q) =1\)  and
\( h_2(pq) =2\)  (cf. [2]}, [3]}).
So Lemma REF  and Kuroda's class number formula (cf. [6]}) imply that
\(h_2(k^{\prime })=\frac{1}{4}q(k^{\prime })h_2(p)h_2(q)h_2(pq)=1\) .
Since there are only two primes
which ramify in \(K_{pq}/k^{\prime }\) , then by ambiguous class number formula (cf. [7]}) the rank of the 2-class group of \(K_{pq}\)  equals
\(2-1-e=1-e\) , where \(e\)  is an integer defined by
\([E_{k^{\prime }}: E_{k^{\prime }}\cap N_{K_{pq}/k^{\prime }}(K_{pq})]=2^e\) . We infer that the rank of the 2-class group of \(K_{pq}\)  can not be equal to 2.
On the other hand, note that \(L_{pq}\) , \(F_{pq}\)  and \(K_{pq}\)  are the three unramified quadratic extensions of \(k_{pq}=\mathbb {Q}(\sqrt{-2}, \sqrt{pq})\)  that having a 2-class group of type \((2, 2)\)  (cf. [8]}). Then, by Remark REF , the 2-class group of \(F_{pq}\)  is of type \((2, 2)\)  and that of \(K_{pq}\)  is cyclic. From [8]}, we deduce that \(q(K_{pq})=4\) . Hence, class number formula (cf. [10]}) implies that
\(h_2(K_{pq})=\frac{1}{2^5}q(K_{pq})h_2(p)h_2(q)h_2(pq)h_2(-2p)h_2(-2q)h_2(-2pq)h_2(-2)=2h_2(-2q).\) 
So the result for this case.
Suppose now that \(p\)  and \(q\)  satisfy conditions \((\ref {cond 2})\) . Using [8]} and [2]}, [3]}, [6]}, we similarly show that \(h_2(K_{pq})=2\cdot h_2(-2q)\) . Thus, \(h_2(K_{pq})\)  is divisible by 8. Thus as above its 2-class group can not be of type \((2, 2)\) , which completes the proof.
